Question: Evaluate the definite integral. $\int^{25}_{0}\left(-6\sqrt{x}\right)\,dx = $
Explanation: First, use the power rule: $\begin{aligned}\int^{25}_{0}\left(-6\sqrt{x}\right)\,dx ~&=~\int^{25}_{0}\left(-6x^{\frac12}\right)\,dx \\&=(-4x^\frac32)\Bigg|^{25}_{0}\end{aligned}$ Second, plug in the limits of integration: $(-4\cdot{25}^{\frac32})-(-4\cdot0^{\frac32}) = -500+0 = -500$. The answer: $\int^{25}_{0}\left(-6\sqrt{x}\right)\,dx ~=~-500$